For the reaction $A \rightleftharpoons B$,find the value of $log_{10}K$. Given: $\Delta_rH^o_{298\,K} = -54.07\, kJ\, mol^{-1}$,$\Delta_rS^o_{298\,K} = 10\, J\, K^{-1}\, mol^{-1}$,$R = 8.314\, J\, K^{-1}\, mol^{-1}$,$2.303 \times 8.314 \times 298 = 5705$.

Assuming ideal behaviour,the magnitude of $\log \, K$ for the following reaction at $25^{\circ} C$ is $x \times 10^{-1}$. The value of $x$ is $......$. (Integer answer)
$3 HC \equiv CH_{(g)} \rightleftharpoons C_6H_{6(\ell)}$
[Given: $\Delta_f G^{\circ}(HC \equiv CH) = -2.04 \times 10^5 \, J \, mol^{-1}$
$\Delta_f G^{\circ}(C_6H_6) = -1.24 \times 10^5 \, J \, mol^{-1}; R = 8.314 \, J \, K^{-1} \, mol^{-1}$]

Find out the value of the equilibrium constant for the following reaction at $298 \, K$.
$2 NH_{3(g)} + CO_{2(g)} \leftrightharpoons NH_{2}CONH_{2(aq)} + H_{2}O_{(l)}$
Standard Gibbs energy change,$\Delta_{r} G^{\ominus}$ at the given temperature is $-13.6 \, kJ \, mol^{-1}$.

For the reaction $CH_{4(g)} + H_{2(g)} \longrightarrow C_2H_{6(g)}$,$K_p = 3.356 \times 10^{17}$,calculate $\Delta G^{\circ}$ for the reaction at $298 \ K$.

Hydrolysis of sucrose is given by the following reaction:
$\text{Sucrose} + H_{2}O \rightleftharpoons \text{Glucose} + \text{Fructose}$
If the equilibrium constant $(K_{c})$ is $2 \times 10^{13}$ at $300 \ K$,the value of $\Delta_{r}G^{\Theta}$ at the same temperature will be:

The equilibrium constant of a reaction is $0.008$ at $298 \ K$. The standard free energy change of the reaction at the same temperature is